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Homework Help: Thermodynamics quiz using ideal gas law

  1. May 8, 2009 #1
    1. The problem statement, all variables and given/known data


    Q: Air initially at 300kpa is contained inside a piston-cylinder device with a volume of 1m^3. The cylinder is equiped with a check valve allowing air to escape when the air pressure reaches 200kPa. An extrernal force pushes the frictionless piston slowly until the final volume reaches 0.1m^3. Durring this entire process, the air temp is maintained at 300K by heat transfer. To determine the heat transfer answer the following questions. Air is assumed to be an ideal gas with constant specefic heats, cp=1 kJ/kg-K, cv=0.713 kJ/kg-K.

    (a) skecth (p-V) (T-V) diagrams [no problem I got those]
    (b) find the heat transfer amount by the time when the valve opens so...Q1-2?
    (c)Write down the energy conservation equation after the valve opens [I got this one too]
    (d)Fing the inital and final mass of air
    (e)Find the heat transfer amount after the valve opens until the final state so...Q2-3?

    2. Relevant equations

    Q-W=0 ===> Q=W
    b/c it is frictionless it is an isotropic process

    3. The attempt at a solution
    (b)for Q1-2 I tried
    Q1-2=mT(s2-s1)=mT(Cpln (T2/T1)-Rln(P2/P1)) = -69.29 ... since s1=s2 I'm unsure if this equation is correct
    I also tried Q1-2=mP(V2-V1) ... however I have 2 different Ps
    So I also tried Q=W={the intergral from 1-2}pdv = m{the intergral from 1-2}C/V^k

    (d)m=PV/RT => m1=1.161 m3=0.232 does m1=m2? ... I alos found? v2= (RT2)/P2=.4305?

    (e) Q2-3=m2P(V3-V2) ... I think this equation is correct but I think I'm missing something....
  2. jcsd
  3. May 9, 2009 #2
    I don't quite understand this question. Is the valve open from the beginning? And if so, then why does it matter if we start with 300kPa, since it will drop to 200kPa instantly (other than to compare initial mass). Also when does the external force start? The question just seems pretty vague.
  4. May 9, 2009 #3
    the valve is closed until state 2 ... so the system goes from a closed to an open
  5. May 9, 2009 #4
    Hmmm... I see it in the variables now, but the problem could have worded it clearer.

    (b) is just an ordinary isothermal process where the work done to the ideal gas system is the heat lost.

    And P2 isn't 200kPa, otherwise process 2->3 would do nothing.

    (d) Can you show more work, it looks like you have the right idea but I can't make it out what you are doing.
  6. May 9, 2009 #5
    The question says:
    valve opens
    Final State

    2-3 is an open system so I believe the enery balance is:
  7. May 9, 2009 #6
    Ok, why do you say "Air initially at 300kpa is contained inside a piston-cylinder device" and you also say P1=100kPa?
  8. May 9, 2009 #7
    the inital state is 300K and 100Kpa .. sry for the confusion
  9. May 9, 2009 #8
    Ok, now the problem makes more sense. That confused me a lot.

    Start with (b). You know it is an isothermal process. You just need to find V2 where P2 = 200kPa for an ideal gas. Then you can get Q using Q = nrT *ln(V2/V1).
  10. May 9, 2009 #9
    okay so ...Q=n(.287)(300K)*ln(.4305/1)
    I got V2=.4305 by V2=RT2/P2
    but what is n
    is n interchangable with k if so k=cp/cv=1.4
    using k=n Q=-101.49 wich makes me think k(not)=n
  11. May 9, 2009 #10
    oh wait n= m/M but how do I find out what M=?
  12. May 9, 2009 #11
    1.161/29 ??? I don't think that is right
  13. May 9, 2009 #12
    Use the ideal gas law, nR = PV/T. You know P,V, and T.

    Also to solve for V2, you know P1*V1 = nRT = P2*V2, so V2 = (P1/P2)*V1.

    The reason you did it wrong before, was because you used Cp-Cv = R. Which is correct, but the units were inaccurate. To use n*R*T=PV you need the units of R = J/(K*mol). But your R is in units R = kJ/(K*kg). So to get the proper units you would use the molar mass of the gas and convert kJ to J.
    Last edited: May 9, 2009
  14. May 9, 2009 #13
    what about Q={intergral} of P r.t.m. => Q12=RTVMl*ln(P2/P1)
  15. May 9, 2009 #14
    that =1.161 PV/T=.334 => .334/R => .334/.287 =1.161=m1
  16. May 9, 2009 #15
    If you want to use Cp-Cv = R, then be sure to convert kJ to J, and use the following equations...

    PV = mRT

    Where 'm' is now the mass of the gas.

    Also 1->2 is an isothermal process. You can easily solve for the heat gained in this process by just knowing the volumes. Or you can rewrite it just knowing the pressures.
  17. May 9, 2009 #16
    How about W12=Q12=mT(Cp*ln(T2/T1)-R*ln(P2/P1))=69.29
    using 214.07 as u
    and 300.19 as h
  18. May 9, 2009 #17
    Your first equation is correct because it was what I said before, but the sign is wrong since heat is lost going from 1 to 2. Also where are your units on 69.29?

    And the last equation I don't know what you are using. What is u and h? And is m the mass?

    Also, could you show your work for (c) and (d).
    Last edited: May 9, 2009
  19. May 9, 2009 #18
    I ment to put -62.29kJ
    I know I need to start w/ an open system mass and energy balance but, I'm not sure where to go from there
  20. May 9, 2009 #19
    (d) m1=PV/RT= 100kPa*1m^3/.287*300K =1.161kg
    m3=PV/RT= 200kPa*0.1m^3/.287*300K =.232kg
    I know the above is correct but,
    (c) I'm unsure of the energy balance
    once I realized my original ideas were inncorrect I don't know what the energy balance is
  21. May 9, 2009 #20
    For (c) you should get something like dE = dQ - dW + dH

    Where dH = change in enthalpy of gas ( for ideal gas dH = Cp*T*dm )

    I might be wrong though, so double check your book.
    Last edited: May 9, 2009
  22. May 9, 2009 #21
    okay..that makes since the answer I got for my previous equation for Q23 matches the answer my prof. gave me... So, thank you
    I posted another question I have been having problems with as well
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